Advertisements
Advertisements
Question
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
Advertisements
Solution
L.H.S. = \[\boxed{\text{cos}^2θ · (1 + \text{tan}^2θ)}\]
= cos2θ × \[\boxed{\text{sec}^2θ}\] ...[1 + tan2θ = \[\boxed{\text{sec}^2θ}\]]
= (cos θ × \[\boxed{\text{sec} θ}\])2
= 12
= 1
= R.H.S.
APPEARS IN
RELATED QUESTIONS
If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 – 1) = 2p
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
Prove the following identities:
`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.
Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.
Prove the following identities.
tan4 θ + tan2 θ = sec4 θ – sec2 θ
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
